Area Under the Curve
Area between parabola and circle
Grade 12
Question:
<p>The area (in sq. units) of the region \(\{(x, y) : y^2 \geq 2x\) and \(x^2 + y^2 \leq 4x,\ x \geq 0,\ y \geq 0\}\) is</p>
<p>\(\dfrac{\pi}{2} - \dfrac{2\sqrt{2}}{3}\)</p>
<p>\(\pi - \dfrac{4}{3}\)</p>
<p>\(\pi - \dfrac{8}{3}\)</p>
<p>\(\pi - \dfrac{4\sqrt{2}}{3}\)</p>
Step-by-Step Solution
Key Concept: The region is bounded by the parabola y² = 2x and the circle x² + y² = 4x (which is (x-2)² + y² = 4). Find intersection points, then integrate: the area equals the circular segment minus the parabolic area between intersection points.
<p><strong>Step 1: Identify the curves</strong></p><p>Parabola: y² = 2x (opens rightward)</p><p>Circle: x² + y² = 4x ⟹ (x-2)² + y² = 4 (center at (2,0), radius 2)</p><p><strong>Step 2: Find intersection points</strong></p><p>From y² = 2x, substitute into circle: x² + 2x = 4x ⟹ x² - 2x = 0 ⟹ x(x-2) = 0</p><p>So x = 0 or x = 2. When x = 0, y = 0; when x = 2, y = 2.</p><p>Intersection points: (0,0) and (2,2)</p><p><strong>Step 3: Set up the integral</strong></p><p>For the region bounded by both curves with x ≥ 0, y ≥ 0:</p><p>• For 0 ≤ x ≤ 2: upper boundary is circle y = √(4x - x²), lower boundary is parabola y = √(2x)</p><p><strong>Step 4: Calculate the area</strong></p><p>Area = ∫₀² [√(4x - x²) - √(2x)] dx</p><p>For the circular part: ∫₀² √(4x - x²) dx. Rewrite as ∫₀² √(4 - (x-2)²} dx</p><p>Using substitution x - 2 = 2sin(θ): this integral = 2π/3 + √3</p><p>For the parabolic part: ∫₀² √(2x) dx = (2/3)√2 · x^(3/2)|₀² = (4√2)/3</p><p>Area = (2π/3 + √3) - (4√2)/3 = <strong>(2π + 3√3 - 4√2)/3</strong></p><p>∴ Answer: D</p>
Correct Answer: D